I need help finding the Inverse Fourier transform of: $$F(k) = \frac1{ k^2 + a^2 },~ a>0$$
Here is what I have so far:
Singular points at $k^2 = a^2$, namely, at $k = \pm ia$. The inverse transform by definition is
$$\;f(x) = \frac{1}{2\pi}\int^{\infty}_{-\infty}F(k)e^{ikx}dk~.$$
Therefore,
\begin{align}
f(x) &= \frac1{2\pi}\int^{\infty}_{-\infty} \frac{e^{ikx}}{k^2+a^2}dk \\
&= \frac1{2\pi} \left[ \int^{0}_{-\infty}\frac{e^{ikx}}{k^2+a^2}dk + \int^{\infty}_{0}\frac{e^{ikx}}{k^2+a^2}dk \right] \\
&= \frac{1}{2\pi}2\pi i \cdot \sum(\operatorname{Res})
\end{align}
Where do I go from here?
I will assume that $x$ is real. You need to pick either the upper or the lower semicircle, based on the sign of $x$. Let $k=u+iv$ (i.e. $\Re(k)=:u$, $\Im(k)=:v$), then we have that $$e^{ikx}=e^{i(u+iv)x}=e^{iux}e^{-vx}$$ We want $e^{ikx} \to 0$. The first part just oscillates, so we don't need to care about it. The second part goes to $0$ if $vx \to \infty$, i.e. we need $vx$ to be positive. So which semicircle do we need if $x>0$, and which one if $x<0$?